classical and modular approaches to exponential diophantine … · classical and modular approaches...

Compositio Math. 142 (2006) 31–62doi:10.1112/S0010437X05001739

Classical and modular approaches to

exponential Diophantine equations

II. The Lebesgue–Nagell equation

Yann Bugeaud, Maurice Mignotte and Samir Siksek

Abstract

This is the second in a series of papers where we combine the classical approach to expo-nential Diophantine equations (linear forms in logarithms, Thue equations, etc.) with amodular approach based on some of the ideas of the proof of Fermat’s Last Theorem. Inthis paper we use a general and powerful new lower bound for linear forms in three log-arithms, together with a combination of classical, elementary and substantially improvedmodular methods to solve completely the Lebesgue–Nagell equation x2 + D = yn, x, yintegers, n � 3, for D in the range 1 � D � 100.

1. Introduction

Arguably, the two most celebrated achievements of the 20th century in the field of Diophantineequations have been Baker’s theory of linear forms in logarithms, and Wiles’ proof of Fermat’s LastTheorem. We call Baker’s approach to Diophantine equations the ‘classical approach’. The proofof Fermat’s Last Theorem is based on what we term the ‘modular approach’. The proponents ofthe classical approach are too many to mention; the modular approach is still in its infancy, butamong the early contributers let us just mention Frey, Serre, Ribet, Darmon, Merel, Kraus, Bennett,Skinner, Ivorra, etc.

The motivation for our series of papers, of which this is the second, is that neither approach(on its own and as it stands at the moment) is powerful enough to resolve unconditionally many ofthe outstanding exponential Diophantine equations. Our thesis is that one should, where possible,attack exponential Diophantine equations by a combination of classical and modular approaches.The precise aims of this series were formulated in our first paper [BMS] as follows.

(I) To present theoretical improvements to some aspects of the classical approach.

(II) To show how local information obtained through the modular approach can be used to reducethe size of the bounds, both for exponents and for variables, of solutions to exponentialDiophantine equations.

(III) To show how local information obtained through the modular approach can be pieced togetherto provide a proof that there are no missing solutions less than the bounds obtained in (I),(II).

(IV) To solve various famous exponential Diophantine equations.

Received 12 May 2004, accepted in final form 18 April 2005.2000 Mathematics Subject Classification 11D61, 11J86 (primary), 11D59, 11Y50 (secondary).Keywords: Diophantine equations, Ramanujan–Nagell, Frey curves, level-lowering, linear forms in logarithms, Thueequations.

S. Siksek’s work is funded by a grant from Sultan Qaboos University (IG/SCI/DOMS/02/06).This journal is c© Foundation Compositio Mathematica 2006.

https://www.cambridge.org/core/terms. https://doi.org/10.1112/S0010437X05001739Downloaded from https://www.cambridge.org/core. IP address: 54.39.106.173, on 19 Oct 2020 at 07:07:56, subject to the Cambridge Core terms of use, available at

http://www.compositio.nl

http://www.ams.org/msc/

http://www.compositio.nl

https://www.cambridge.org/core/terms

https://doi.org/10.1112/S0010437X05001739

https://www.cambridge.org/core

Y. Bugeaud, M. Mignotte and S. Siksek

In [BMS] we gave a new lower bound for linear forms in three logarithms and used a combinationof classical and modular methods to determine all the perfect powers in the Fibonacci and Lucassequences. In the present paper, we apply a more general and powerful lower bound for linear formsin three logarithms due to Mignotte [Mig], together with a combination of elementary, classical andsubstantially improved modular methods to study the following exponential Diophantine equation:

x2 + D = yn, x, y integers, n � 3. (1)

Here, D denotes a non-zero integer. We have chosen to name this equation the Lebesgue–Nagellequation; the reason for the name Lebesgue–Nagell is given in § 2, together with some historicalremarks. However, for now we mention that the equation has previously been solved for 81 valuesof D in the range 1 � D � 100, using elementary, classical and modular methods; the remainingvalues are apparently beyond these methods as they stand. We prove the following theorem.

Theorem 1. All solutions to (1) with D in the range

1 � D � 100 (2)

are given in the table in Appendix A. In particular, the only integer solutions (x, y, n) to thegeneralised Ramanujan–Nagell equation

x2 + 7 = yn, n � 3,

satisfy |x| = 1, 3, 5, 11, 181.

We choose to give a complete proof of Theorem 1, rather than treating the 19 remaining valuesof D in the range (2).

It is noted that the solutions for even n can be deduced quickly, for then D is expressible as adifference of squares. It is therefore sufficient to solve the equation

x2 + D = yp, x, y integers, p � 3 is prime; (3)

the solutions to (1) can then be recovered from the solutions to (3).We give three modular methods for attacking (3). Two are refinements of known methods and

a third that is completely new. Using a computer program based on these modular methods, wecan show, for any D in the above range, that the exponent p is large (showing that p > 109 isquite practical). Our modular approach also yields the following rather surprising result: eithereach prime factor of y divides 2D or y > (

√p − 1)2. We are then able to deduce not only that p is

large, but also that y is large. This information helps us to reduce the size of the upper bound onp obtained from the lower bound for the linear forms in three logarithms, making the computationmuch more practical. This idea of using the modular approach to force lower bounds for solutionsof Diophantine equations was used previously, for instance by Bennett [Ben04]. Our total computertime for the computations in this paper is roughly 206 days on various workstations (the precisedetails are given in due course).

Using our approach should make it possible to solve (1) for any D, with |D| not too large, that isnot of the form D = −a2 ± 1; if D is of this form then (1) has a solution (x, y) = (a,±1) for all oddvalues of the exponent n, and the modular methods we explain later are not very successful in thissituation. To deal with this case requires further considerations, which we leave for another paper.Note, however, that the case D = 1 turns out to be quite easy and was solved in 1850 by Lebesgue[Leb50]. Furthermore, (1) has been solved for some negative values of D of the form D = −a2 ± 1,including D = −1 and D = −8 (see, for example, [Ivo03, Sik03]). However, proving that the onlyinteger solutions to x2 − 2 = yn with n � 3 satisfy |x| = 1 remains a challenging open question. Forsome modest progress on this question, due to the authors, see [Sik].

32



https://doi.org/10.1112/S0010437X05001739


Classical and modular approaches

2. On the history of the Lebesgue–Nagell equation

Equation (1) has a long and glorious history and there are literally hundreds of papers devoted tospecial cases of this equation. Most of these are concerned with (1) either for special values of n orspecial values of y. For example, for D = 2 and n = 3, Fermat asserted that he had shown that theonly solutions are given by x = 5, y = 3; a proof was given by Euler [Eul70]. Equation (1) withn = 3 is the intensively studied Mordell equation (see [GPZ98] for a modern approach).

Another notable special case is the generalized Ramanujan–Nagell equation

x2 + D = kn, (4)

where D and k are given integers. This is an extension of the Ramanujan–Nagell equation x2+7 = 2n,proposed by Ramanujan [Ram13] in 1913 and first solved by Nagell [Nag48] in 1948 (see also thecollected papers of Nagell [Nag02]). This equation has exactly five solutions with x � 1 (see [Mig84]for a very simple proof) and is, in this respect, singular: indeed, Bugeaud and Shorey [BS01] estab-lished that (4) with D positive and k a prime number not dividing D has at most two solutions inpositive integers x, n, except for (D, k) = (7, 2). They also listed all of the pairs (D, k) as abovefor which (4) has exactly two solutions. Much earlier, Apery, [Ape60a, Ape60b] proved by p-adicarguments that x2 + D = kn, with k prime, has at most two positive integer solutions except if(D, k) = (7, 2). We direct the reader to [BS01] for further results and references.

Returning to (1), the first result for general y, n seems to be the proof in 1850 by Lebesgue[Leb50] that there are no non-trivial solutions for D = 1. The next cases to be solved were D = 3, 5by Nagell [Nag48] in 1923. It is for this reason that we call (1) the Lebesgue–Nagell equation. Thecase with D = −1 is particularly noteworthy: a solution was sought for many years as a special caseof the Catalan conjecture. This case was finally settled by Ko [Ko65] in 1965.

The history of the Lebesgue–Nagell equation is meticulously documented in an important articleby Cohn [Coh93b] and so we are saved the trouble of compiling an exhaustive survey. In particular,Cohn refines the earlier elementary approaches of various authors (especially of Ljunggren [Lju63,Lju64]) and completes the solution for 77 values of D in the range 1 � D � 100. The solution forthe cases D = 74, 86 was completed by Mignotte and de Weger [MW96] (indeed, Cohn solved thesetwo equations of type (3) except for p = 5, in which case difficulties occur as the class numbersof the corresponding imaginary quadratic fields are divisible by 5). Bennett and Skinner [BS04,Proposition 8.5] applied the modular approach to solve the cases D = 55 and 95. The 19 remainingvalues

7, 15, 18, 23, 25, 28, 31, 39, 45, 47, 60, 63, 71, 72, 79, 87, 92, 99, 100, (5)

are clearly beyond the scope of Cohn’s elementary method, although Cohn’s method can still givenon-trivial information even in these cases and is revisited in § 5. Moreover, as far as we can see,the modular method used by Bennett and Skinner (which is what we call Method I) is not capableof handling these values on its own, even though it still gives useful information in most cases.

Cohn [Coh93b], also makes a challenge of proving that the only solutions to the equation

x2 + 7 = yn

have |x| = 1, 3, 5, 11, 181. This challenge was taken up by Lesage [Les98] who proved, by classicalarguments, that if x > 181 then 5000 < n < 6.6×1015 and also by Siksek and Cremona [SC03] whoused the modular approach to show that there are no further solutions for n � 108 (consequently,n must be prime); they also suggested that an improvement to lower bounds in linear forms inthree logarithms may finally settle the problem. With the benefit of hindsight, we know that theywere almost, although not entirely, correct. The substantial improvement to lower bounds in linearforms in three logarithms used here was certainly needed. However, for this lower bound to be

33



https://doi.org/10.1112/S0010437X05001739



even more effective, a further insight obtained from the modular approach was also needed: namelythat y is large as indicated in the introduction; note further that Lesage proved that y > 109 byclassical arguments (linear forms in two 2-adic logarithms from [BL96]) and some heavy computerverification.

3. Reduction to Thue equations

Our main methods for attacking (3) are linear forms in logarithms (to bound p) and the modularapproach, although for some small values of p it is necessary to reduce the equation to a family ofThue equations. The method for reducing (3) to Thue equations is well known. We do, however,feel compelled to give a succinct recipe for this, in order to set up notation that is needed later.

It is appropriate to point out that there are other approaches that could be used to solve (3)for small p. For p = 3 we can view the problem as that of finding integral points on an ellipticcurve, a problem that is aptly dealt with in the literature (see [Sma98, GPZ98]). For p � 5, theequation x2 + D = yp defines a curve of genus � 2; one can sometimes determine all rational pointson this curve using the method of Chabauty [CF96], although this would also require computingthe Mordell–Weil group of the Jacobian (see [PS97, Sch95a, Sto98, Sto01, Sto02]).

We do not assume in this section that D is necessarily in the range (2), merely that −D is nota square. We write (here and throughout the paper)

D = D21D2, D1, D2 are integers, D2 square-free.

Let L = Q(√−D2) and O be its ring of integers. Throughout the present paper, we denote the

conjugate of an element α (respectively of an ideal a) by α (respectively by a).Let p1, . . . , pr be the prime ideals of O dividing 2D. Let A be the set of integral ideals a of O

such that:

• a = pa11 · · · par

r , with 0 � ai < p;• the gcd(a, a) divides 2D1

√−D2;

• the ideal aa is a perfect pth power.

If (x, y) is a solution to (3), then one effortlessly sees that

(x + D1

√−D2)O = abp

for some a ∈ A and some integral ideal b.Now let b1, . . . , bh be integral ideals forming a complete set of representatives for the ideal class

group of O. Thus, bbi is a principal ideal for some i and so bbi = β′O for some β′ ∈ O. The fractionalideal ab

−pi is easily seen to be also principal. The ideal b is unknown, but the ideals, a, b1, . . . , bh are

known. We may certainly determine which of the fractional ideals ab−pi are principal. Let Γ′ be a

set containing one generator γ′ for every principal ideal of the form ab−pi (a ∈ A and 1 � i � h). It

is noted that the elements of Γ′ are not necessarily integral, but we know that if (x, y) is a solutionto (3) then

(x + D1

√−D2)O = γ′β′pO,

for some γ′ ∈ Γ′ and some β′ ∈ O. Finally, define Γ as follows:

Γ =

Γ′, if D2 > 0, D2 �= 3, or if D2 = 3 and p �= 3,

Γ′ ∪ ζΓ′ ∪ ζ−1Γ′, if D2 = p = 3, where ζ = (1 +√−3)/2,⋃

j

εjΓ′, if D2 < 0, where j runs over −(p − 1)/2, . . . , (p − 1)/2,

where if D2 < 0 (and so L is real) we write ε for the fundamental unit.

34



https://doi.org/10.1112/S0010437X05001739



We quickly deduce the following.

Proposition 3.1. With notation as above, if (x, y) is a solution to (3) then there exist γ ∈ Γ andβ ∈ O such that

x + D1

√−D2 = γβp.

Thus if we let 1, ω be an integral basis for O then for some γ ∈ Γ,

x = 12(γ(U + V ω)p + γ(U + V ω)p)

for some integral solution (U, V ) to the Thue equation

12√−D2

(γ(U + V ω)p − γ(U + V ω)p) = D1.

3.1 Results IIf q is a prime we denote by vq : Z → Z�0 ∪ {∞} the normalized q-adic valuation.

We now eliminate all cases where it is inconvenient to carry out level-lowering.

Lemma 3.2. Suppose that 1 � D � 100. Suppose that (x, y, p) is a solution to (3) that is missingfrom our table in Appendix A. Then p satisfies the following conditions:

p � 7,p � vq(D) + 1, for all primes q,

p � v2(D) + 7, if v2(D) is even.

(6)

Proof. It is clear that for any particular D there are only a handful of primes p violating any ofthese conditions. We wrote a pari/gp [BBBCO] program that solved all (3) for p violating (6): theprogram first reduces each such equation to a family of Thue equations as in Proposition 3.1 above.These are then solved using the built-in pari/gp function for solving Thue equations (this is animplementation of the method of Bilu and Hanrot [BH96]).

It is perhaps worthwhile to record here two tricks that helped us in this step. First, in writingdown the set Γ appearing in Proposition 3.1 we needed a set of integral ideals b1, . . . , bh representingthe ideal class group of the quadratic field L. Both pari/gp and MAGMA [BCP97] have built-infunctions that amount to homomorphisms from the ideal class group as an abstract group, to theset of fractional ideals, and these can be used to construct the required set b1, . . . , bh. We havefound, however, that we get much simpler Thue equations if we search for the smallest prime idealrepresenting each non-trivial ideal class, and of course taking 1O to represent the trivial ideal class.

To introduce the second trick, we recall that when one is faced with a Thue equation

a0Up + a1U

p−1V + · · · + apVp = b

it is usual to multiply throughout by ap−10 and make the substitution U ′ = a0U , thus obtaining a

monic polynomial on the left-hand side. When a0 is large, this greatly complicates the equation.The second trick is to first search for a unimodular substitution, which makes the leading coefficienta0 small.

After optimizing our program, we were able to complete the proof of Lemma 3.2 in about 22minutes on a 1050 MHz UltraSPARC III computer.

4. Removing common factors

It is desirable when applying the modular approach to (3) to remove the possible common factors ofthe three terms in the equation. This desire leads to a subdivision of cases according to the possible

35



https://doi.org/10.1112/S0010437X05001739



common factors, as seen in the following elementary lemma. Here and elsewhere, for a non-zerointeger a, the product of the distinct prime divisors of a is called the radical of a, and denoted byrad(a), in particular rad(±1) = 1. Furthermore, ( ··) stands for the Kronecker symbol.

Lemma 4.1. Suppose that (x, y, p) is a solution to (3) such that y �= 0 and p satisfies the condi-tion (6). Then there are integers d1, d2 such that the following conditions are satisfied:

(i) d1 > 0;(ii) D = d2

1d2;(iii) gcd(d1, d2) = 1;(iv) for all odd primes q|d1 we have (−d2

q ) = 1;(v) if 2|d1 then d2 ≡ 7 (mod 8).

Moreover, there are integers s, t such that

x = d1t, y = rad(d1)s,

and

t2 + d2 = esp, gcd(t, d2) = 1, s �= 0, (7)where

e =∏

q primeq|d1

qp−2vq(d1) and rad(e) = rad(d1). (8)

Proof. Suppose that (x, y, p) is a solution to (3) such that y �= 0 and condition (6) is satisfied. It isstraightforward to see that condition (6) forces gcd(x2,D) to be a square, say d2

1 with d1 > 0. Wecan therefore write x = d1t and D = d2

1d2 for some integers t and d2. Moreover, because

d21 = gcd(x2,D) = gcd(d2

1t2, d2

1d2) = d21 gcd(t2, d2),

we see that gcd(t, d2) = 1. Removing the common factors from x2 + D = yp we obtain t2 + d2 = esp

where e is given by (8). The integrality of e follows from the condition (6), and so does the equalityof the radicals rad(e) = rad(d1). Note that (iii) follows from this equality of the radicals and the factthat t, d2 are coprime. We have thus proven (i), (ii), (iii) and it is now easy to deduce (iv) and (v).Finally, the condition s �= 0 follows from the condition y �= 0.

Definition. Suppose that D is a non-zero integer and (x, y, p) is a solution to (3) with y �= 0 andp satisfying (6). Let d1, d2 be as in the above lemma and its proof (thus d1 > 0, gcd(x,D) = d2

1 andd2 = D/d2

1). We call the pair (d1, d2) the signature of the solution (x, y, p). We call the pair (t, s)the simplification of (x, y) (or (t, s, p) the simplification of (x, y, p)).

In this terminology, Lemma 4.1 associates with any D a finite set of possible signatures (d1, d2)for the solutions (x, y, p) of (3) satisfying (6) and y �= 0. To solve (3) it is sufficient to solve it underthe assumption that the solution’s signature is (d1, d2) for each possible signature.

Example 1. For example, if D = 25, there are two possible signatures satisfying the conditions ofLemma 4.1; these are (d1, d2) = (1, 25) or (5, 1). If (d1, d2) = (1, 25), then x = t, y = s and we mustsolve the equation

t2 + 25 = sp, 5 � t,

already solved by Cohn. However, if (d1, d2) = (5, 1), then x = 5t, y = 5s, and we must solve theequation

t2 + 1 = 5p−2sp,

not solved by Cohn. In either case it is noted that the three terms of the resulting equation arerelatively coprime, which is important to apply the modular approach.

36



https://doi.org/10.1112/S0010437X05001739



5. A simplification of Cohn

We will soon apply our modular machinery to (3) with D in the range (2). Before doing thisit is helpful to introduce a simplification due to Cohn that will drastically reduce the amountof computation needed later. All the arguments presented in this section are found in Cohn’spapers [Coh93b, Coh03]. Cohn, however, assumed that D �≡ 7 (mod 8); the result below is notsubject to this restriction.

Proposition 5.1. Let D = D21D2 where D2 is square-free and D2 > 0. Suppose that (x, y, p) is a

solution to (3) with p satisfying (6) and let (d1, d2) be the signature of this solution. Then:

(i) d1 > 1; or

(ii) D ≡ 7 (mod 8) and 2|y; or

(iii) p divides the class number h of the quadratic field Q(√−D2); or

(iv) y = a2 + D2b2 for some integers a and b such that b|D1, b �= ±D1,

p|(D21 − b2) and

12√−D2

[(U + b√

−D2)p − (U − b√

−D2)p] = D1; or

(v) D = 1, (x, y) = (0, 1); or

(vi) D2 ≡ 3 (mod 4) and y = (a2 + D2b2)/4 for some odd integers a and b such that b|D1,

p|(4D21 − b2) and a is a solution of the equation

12√−D2

[(U + b√

−D2)p − (U − b√

−D2)p] = 2pD1.

Proof. We only give a brief sketch. Suppose that (i), (ii), (iii) are false. Then (x + D1

√−D2) = αp

for some α in the ring of integers of Q(√−D2). There are two possibilities. The first is that α =

a + b√−D2 for some integers a and b. By equating the imaginary parts we deduce all of (iv) if

b �= ±D1. Thus, suppose that b = ±D1. Letting β = a − b√−D2 we see that

αp − βp

α − β= ±1.

If α/β is not a root of unity, then the left-hand side is the pth term of a Lucas sequence (with p � 7)and a deep theorem of Bilu et al. [BHV01] on primitive divisors of Lucas and Lehmer sequencesimmediately gives a contradiction. Thus α/β is a root of unity, i.e. α/β = ±1, ±i, or (±1±

√−3)/2.

Each case turns out to be impossible, except for α = −β, which together with b = ±D1 implies (v).The second possibility for α is that α = (a + b

√−D2)/2 with a, b odd integers (and −D2 ≡ 1

(mod 4)). Now (vi) follows quickly by equating the imaginary parts of (x + D1

√−D2) = αp.

5.1 Results IICorollary 5.2. Suppose that D belongs to our range (2) and (x, y, p) is a solution to (3) with psatisfying the condition (6). If the solution (x, y, p) is missing from the table in Appendix A, theneither D ≡ 7 (mod 8) and 2|y or d1 > 1, where (d1, d2) is the signature of the solution.

Proof. We apply Proposition 5.1. Using a short MAGMA program we listed all solutions arising frompossibilities (iv)–(vi) of Proposition 5.1 with 1 � D � 100. The only solutions found in our rangeare (x, y, p) = (0, 1, p) for D = 1 and (x, y, p) = (±8, 2, 7) for D = 64 and these are certainly in thetable in Appendix A.

To prove the corollary we merely have to take care of possibility (iii) of the proposition. For1 � D � 100 and primes p satisfying (6), the only case when p could possibly divide the class numberof Q(

√−D2) is p = 7 and D = 71 (in which case h = 7). We solved the equation x2 + 71 = y7

37



https://doi.org/10.1112/S0010437X05001739



by reducing to Thue equations as in § 3. It took pari/gp about 30 minutes to solve these Thueequations, and we obtained that the only solutions are (x, y) = (±46, 3), again in the table inAppendix A.

6. Level lowering

In this section we apply the modular approach to (7) under suitable, but mild, hypotheses. Ordin-arily, one would have to construct a Frey curve or curves associated with our equation, show thatthe Galois representation is irreducible (under suitable hypotheses) using the results of Mazurand others [Maz78] and modular by the work of Wiles and others [Wil95, TW95, BCDT01], andfinally apply Ribet’s level-lowering theorem [Rib90]. Fortunately we are saved much trouble by theexcellent paper of Bennett and Skinner [BS04], which does all of this for equations of the formAxn + Byn = Cz2; it is noted that (7) is indeed of this form.

Let D be a non-zero integer. We shall apply the modular approach to the Diophantine equation

x2 + D = yp, x2 �D, y �= 0 and p � 3 is prime, (9)or the equivalent equation for the simplification (s, t)

t2 + d2 = esp, t �= ±1, gcd(t, d2) = 1, s �= 0, (10)

under the additional assumption that p satisfies (6). The assumptions made about s, t in (10) arethere to ensure the non-singularity of the Frey curves, and the absence of complex multiplicationwhen we come to apply the modular approach later on. Before going on we note the followinglemma, which in effect states that there is no harm in making these additional assumptions for Din our range (2).

Lemma 6.1. There are no solutions to (3) for D in the range (2) with y = 0, or x2|D, except thoselisted in the table in Appendix A.

Proof. Clearly y �= 0. We produced our list of solutions with x2|D using a short MAGMA program.

Lemma 4.1 associates with each equation of the form (9) finitely many signatures (d1, d2) satis-fying conditions (i)–(v) and corresponding (7). Following Bennett and Skinner [BS04], we associatea Frey curve Et with any potential solution of (10) according to Tables 1–3.

Tables 1–3 are divided into cases (a)–(l). We know that d1, d2 are coprime and, hence, at mostone of them is even. The possibility that d1, d2 are both odd is dealt with in Table 1. In cases(a), (b), a simple modulo 8 argument convinces us that t is odd. However, for cases (c) and (d),where d1 is odd and d2 ≡ 7 (mod 8), the integer t can be either odd or even and we assign differentFrey curves for each possibility. When t is odd (case (d)) we add the assumption that t ≡ 1 (mod 4);this can be achieved by interchanging t with −t if necessary.

Table 2 deals with the possibility of even d1 and Table 3 deals with the possibility of even d2.In both of these cases t is necessarily odd and the congruence condition on t can again be achievedby interchanging t with −t if necessary.

Proposition 6.2. Suppose that D, d1, d2 are non-zero integers that satisfy (i)–(v) of Lemma 4.1.Suppose also that p is a prime satisfying (6) and let e be as defined in (8). Suppose that (t, s) is asolution of (10) and satisfies the supplementary condition (if any) on t in Tables 1–3. Let Et andL be as in these tables and write ρp(Et) for the Galois representation on the p-torsion of Et. Thenthe representation ρp(Et) arises from a cuspidal newform of weight 2 and level N = L rad(D).

Proof. In [BS04], Bennett and Skinner give an exhaustive recipe for Frey curves and level loweringfor equations of the form Axn + Byn = Cz2 under the assumption that the three terms in the

38



https://doi.org/10.1112/S0010437X05001739



Table 1. Frey curves with d1, d2 odd.

Case Condition on d2 Condition on t Frey curve Et L

(a) d2 ≡ 1 (mod 4) Y 2 = X3 + 2tX2 − d2X 25

(b) d2 ≡ 3 (mod 8) Y 2 = X3 + 2tX2 + (t2 + d2)X 25

(c) d2 ≡ 7 (mod 8) t even Y 2 = X3 + 2tX2 + (t2 + d2)X 25

(d) d2 ≡ 7 (mod 8) t ≡ 1 (mod 4) Y 2 + XY = X3 +(

t − 14

)X2 +

(t2 + d2

64

)X 2

Table 2. Frey curves with d1 even, d2 odd.

Case Conditions on t, s, p Frey curve Et L

(e) t ≡ 1 (mod 4) Y 2 + XY = X3 +(

t − 14

)X2 +

(t2 + d2

64

)X 1

Table 3. Frey curves with d1 odd, d2 even.

Case Condition on d2 Condition on t Frey curve Et L

(f) v2(d2) = 1 Y 2 = X3 + 2tX2 − d2X 26

(g) d2 ≡ 4 (mod 16) t ≡ 1 (mod 4) Y 2 = X3 + tX2 − d2

4X 2

(h) d2 ≡ 12 (mod 16) t ≡ 3 (mod 4) Y 2 = X3 + tX2 − d2

4X 22

(i) v2(d2) = 3 t ≡ 1 (mod 4) Y 2 = X3 + tX2 − d2

4X 24

(j) v2(d2) = 4, 5 t ≡ 1 (mod 4) Y 2 = X3 + tX2 − d2

4X 22

(k) v2(d2) = 6 t ≡ 1 (mod 4) Y 2 + XY = X3 +(

t − 14

)X2 − d2

64X 2−1

(l) v2(d2) � 7 t ≡ 1 (mod 4) Y 2 + XY = X3 +(

t − 14

)X2 − d2

64X 1

equation are coprime. After a little relabeling, their results apply to (10) and the lemma followsfrom §§ 2 and 3 of their paper. It is here that we need the assumptions t �= ±1 and s �= 0 madein (10).

It is convenient to indulge in the following abuse of language.

Definition. If (t, s, p) is a solution to (10) and if the representation ρp(Et) arises from a cuspidalnewform f , then we say that solution (t, s, p) arises from the newform f (via the Frey curve Et), orthat the newform f gives rise to the solution (t, s, p). If (t, s, p) is the simplification of (x, y, p) thenwe say that (x, y, p) arises from the newform f . If the newform f is rational and so corresponds toan elliptic curve E, then we also say that the solution (t, s, p) (or (x, y, p)) arises from E.

6.1 A summaryIt may be helpful for the reader to summarize what we have done and where we are going. Givena non-zero integer D we would like to solve (9). We can certainly write down all solutions withy = 0 or with x2|D. We can also solve (at least in principle) all cases where p violates condition (6)by reducing to Thue equations as in § 3. We can thus reduce to (9) and assume that p satisfiescondition (6).

39



https://doi.org/10.1112/S0010437X05001739



Next, we can write down a list of signatures (d1, d2) satisfying conditions (i)–(v) of Lemma 4.1.We reduce the solution of (9) to solving (10) for each signature (d1, d2). Now we associate with thesignature (d1, d2) one or more Frey curves Et and levels N , so that any solution to (10) arises fromsome newform f at level N via the Frey curve Et.

Finally (and this is to come) we must show how to solve (10) under the assumption that thesolution arises from a newform f via a Frey curve Et. If we can do this for each newform f atthe necessary level and Frey curve Et, then we will have completed the solution of our (3).

As we shall see, the assumption that a solution arises from a particular newform is a very strongone, for it imposes congruence conditions on t modulo all but finitely many primes l.

6.2 CongruencesFor an elliptic curve E and a prime of good reduction l we write �E(Fl) for the number of pointson E over the finite field Fl, and let al(E) = l + 1 − �E(Fl).

Lemma 6.3. With notation as above, suppose that the Galois representation ρp(Et) arises from acuspidal newform with Fourier expansion around infinity

f = q +∑n�2

cnqn, (11)

of level N (given by Proposition 6.2) and defined over a number field K/Q. Then there is a placeP of K above p such that for every prime l �2pD we have

al(Et) ≡ cl (mod P), if t2 + d2 �≡ 0 (mod l) (or equivalently l �s),

l + 1 ≡ ±cl (mod P), if t2 + d2 ≡ 0 (mod l) (or equivalently l|s).

Proof. The lemma is standard (see [Ser87, p. 196], [BS04, p. 7], [Kra98, Proposition 5.4], etc.). Theconditions l �2D and l �s together imply that l is a prime of good reduction for Et, whereasthe conditions l �2D and l|s imply that l is a prime of multiplicative reduction.

When the newform f is rational, there is an elliptic curve E defined over Q whose conductor isequal to the level of the newform f such that al(E) = cl for all primes of good reduction l. In thiscase we can be a little more precise than in Lemma 6.3, thanks to a result of Kraus and Oesterle.

Lemma 6.4. With notation as above, suppose that the Galois representation ρp(Et) arises from arational cuspidal newform f corresponding to an elliptic curve E/Q. Then for all primes l �2D wehave

al(Et) ≡ al(E) (mod p), if t2 + d2 �≡ 0 (mod l) (or equivalently l �s),

l + 1 ≡ ±al(E) (mod p), if t2 + d2 ≡ 0 (mod l) (or equivalently l|s).

Proof. This lemma does appear to be a special case of Lemma 6.3; however, we do allow in thislemma the case l = p, which was excluded before. In fact, Lemma 6.3 together with a result ofKraus and Oesterle [KO92, Proposition 3] implies that the representations ρp(Et) and ρp(E) aresemi-simply isomorphic. In this case the result of Kraus and Oesterle also tells us that al(Et) ≡ al(E)(mod p) if the prime l is a prime of good reduction for both curves, and al(Et)al(E) ≡ l+1 (mod p)if l is a prime of good reduction for one of them and a prime of multiplicative reduction for the other.Now, because l �2D we see that l �N , the conductor N of E (which is also the level of the newformf as given by Proposition 6.2). If l|s, then l is a prime of multiplicative reduction for Et and thenal(Et) = ±1. The lemma follows.

40



https://doi.org/10.1112/S0010437X05001739



7. Eliminating exponents: Method I

We now focus on equations of the form (10) where, as always, p satisfies (6). Proposition 6.2 tells usthat if (t, s, p) is a solution to (10), then it arises from a newform of a certain level (or levels) and allof these can be determined. Let us say that these newforms are f1, . . . , fn. Then to solve (10) it issufficient to solve it, for each i, under the assumption that the solution arises from the newform fi.We give three methods for attacking (10) under the assumption that the solution arises from aparticular newform f .

If successful, the first method will prove that (10) has no solutions except possibly for finitelymany exponents p and these are determined by the method. This method is actually quite standard.As far as we know the basic idea is originally due to Serre [Ser87, pp. 203–204]. It is also found inBennett and Skinner [BS04, Proposition 4.3]. We shall, however, give a more careful version thanis found in the literature, thereby maximizing the probability of success.

Proposition 7.1 (Method I). Let D, d1, d2 be a triple of integers satisfying Lemma 4.1(i)–(v).Let f be a newform with Fourier expansion as in (11) having coefficients in the ring of integers ofa number field K, and let NK/Q denote the norm map. If l �2D is prime, let

B′′l (f) = lcm{NK/Q(al(Et) − cl) : t ∈ Fl, t

2 + d2 �≡ 0 (mod l)},

B′l(f) =

B′′l (f), if

(−d2

l

)= −1,

lcm{B′′l (f),NK/Q(l + 1 + cl),NK/Q(l + 1 − cl)}, if

(−d2

l

)= 1,

and

Bl(f) =

{lB′

l(f), if K �= Q,

B′l(f), if K = Q.

If p satisfies condition (6), and if (t, s, p) is a solution to (10) arising from the newform f , then pdivides Bl(f).

Proof. The proposition follows almost immediately from Lemmas 6.3 and 6.4.

Under the assumptions made (in this proposition), Method I eliminates all but finitely manyexponents p, provided of course that Bl(f) is non-zero. Accordingly, we shall say that Method Iis successful if there exists some prime l �2D so that Bl(f) �= 0. There are two situations whereMethod I is guaranteed to succeed.

• If the newform f is not rational. In this case, for infinitely many primes l, the Fourier coefficientcl �∈ Q and so all the differences al(Et) − cl and l + 1 ± cl are certainly non-zero, immediatelyimplying that Bl(f) �= 0.

• Suppose that the newform f is rational and so corresponds to an elliptic curve E definedover Q. Suppose that E has no non-trivial 2-torsion. By the Cebotarev Density Theorem weknow that �E(Fl) is odd for infinitely many primes l. Let l �2D be any such prime. Fromthe models for the Frey curves Et in Tables 1–3 we see that Et has non-trivial 2-torsion,and so l + 1 − al(Et) = �Et(Fl) is even for any value of t ∈ Fl, t2 + d2 �= 0. In this caseal(Et)− cl = al(Et)− al(E) must be odd and cannot be zero. Similarly, the Hasse–Weil bound|cl| � 2

√l implies that l + 1 ± cl �= 0. Thus Bl(f) is non-zero in this case and Method I is

successful.

41



https://doi.org/10.1112/S0010437X05001739



8. Eliminating exponents: Method II

The second method is adapted from the ideas of Kraus [Kra98] (see also [SC03]). It can only beapplied to one prime (exponent) p at a time and, if successful, it does show that there are nosolutions to (10) for that particular exponent.

Let us briefly explain the idea of this second method. Suppose that f is a newform with Fourierexpansion as in (11) and suppose that p � 7 is a prime. We are interested in solutions to (10) arisingfrom f . Choose a small integer n so that l = np + 1 is prime with l �D. Suppose that (t, s) is asolution to (10) arising from f . Working modulo l we see that d2

1t2 + D = yp is either 0 or an nth

root of unity. (Indeed (yp)n = yl−1 ≡ 0 or 1 (mod l).) As n is small we can list all such t in Fl,and compute cl and al(Et) for each t in our list. We may then find that for no t in our list are therelations in Lemma 6.3 satisfied. In this case we have a contradiction and we deduce that there areno solutions to (10) arising from f for the exponent p.

Let us now write this formally. Suppose that p � 7 is a prime number and n an integer suchthat l = np + 1 is also prime and l �D. Define

µn(Fl) = {ζ ∈ F∗l : ζn = 1} and A(n, l) =

{ζ ∈ µn(Fl) :

(ζ − D

l

)= 0 or 1

}.

For each ζ ∈ A(n, l), let δζ be an integer satisfying

δ2ζ ≡ (ζ − D)/d2

1 (mod l).

It is convenient to write al(ζ) for al(Eδζ). We can now give our sufficient condition for the insolubility

of (10) for the given exponent p.

Proposition 8.1 (Method II). Let D, d1, d2 be a triple of integers satisfying Lemma 4.1(i)–(v),and let p � 7 be a prime satisfying condition (6). Let f be a newform with Fourier expansion asin (11) defined over a number field K. Suppose that there exists an integer n � 2 satisfying thefollowing conditions.

(a) The integer l = np + 1 is prime, and l �D.

(b) Either (−d2l ) = −1, or p �NK/Q(4 − c2

l ).(c) For all ζ ∈ A(n, l) we have{

p �NK/Q(al(ζ) − cl), if l ≡ 1 (mod 4),p �NK/Q(al(ζ)2 − c2

l ), if l ≡ 3 (mod 4).

Then (10) does not have any solutions for the given exponent p arising from the newform f .

Proof. Suppose that the hypotheses of the proposition are satisfied and that (t, s) is a solutionto (10).

First we show that t2 + d2 �≡ 0 (mod l). Suppose otherwise. Thus t2 + d2 ≡ 0 (mod l) and sol|s. In this case (−d2

l ) = 1 and from (b) we know that p does not divide NK/Q(4− c2l ). However, by

Lemma 6.3 we know that ±cl ≡ l + 1 ≡ 2 (mod P) for some place P of K above p and we obtaina contradiction showing that t2 + d2 �≡ 0 (mod l).

From (10) and the definition of e in (8), we see the existence of some ζ ∈ A(n, l) such that

d21t

2 + D ≡ ζ (mod l) and t ≡ ±δζ (mod l).

Replacing t by −t in the Frey curve Et has the effect of twisting the curve by −1 (this can be easilyverified for each Frey curve in Tables 1–3). Thus al(ζ) = al(Et) if l ≡ 1 (mod 4) and al(ζ) = ±al(Et)if l ≡ 3 (mod 4). Moreover, by Lemma 6.3, al(Et) ≡ cl (mod P) for some place P of K above p.This clearly contradicts (c). Hence, there is no solution to (10) arising from f for the exponent p.

42



https://doi.org/10.1112/S0010437X05001739



If the newform f is rational and moreover corresponds to an elliptic curve with 2-torsion, then itis possible to strengthen the conclusion of Proposition 8.1 by slightly strengthening the hypotheses.The following variant is far less costly in computational terms as we explain below.

Proposition 8.2 (Method II). Let D, d1, d2 be a triple of integers satisfying Lemma 4.1(i)–(v), andlet p be a prime satisfying condition (6). Let f be a rational newform corresponding to elliptic curveE/Q with 2-torsion. Suppose that there exists an integer n � 2 satisfying the following conditions.

(a) The integer l = np + 1 is prime, l � p2/4 and l �D.

(b) Either (−d2l ) = −1, or al(E)2 �≡ 4 (mod p).

(c) For all ζ ∈ A(n, l) we have {al(ζ) �= al(E), if l ≡ 1 (mod 4),al(ζ) �= ±al(E), if l ≡ 3 (mod 4).

Then (10) does not have any solutions for the given exponent p arising from the newform f .

Proof. Comparing this with Proposition 8.1 we see that it is sufficient to show, under the additionalassumptions, that if al(ζ)2 ≡ al(E)2 (mod p) then al(ζ) = ±al(E), and if al(ζ) ≡ al(E) (mod p)then al(ζ) = al(E).

Suppose that al(ζ)2 ≡ al(E)2 (mod p) (the other case is similar). Hence, al(ζ) ≡ ±al(E)(mod p). Now note that both elliptic curves under consideration here have 2-torsion. Hence, wecan write al(ζ) = 2b1 and al(E) = 2b2 for some integers b1 and b2. Moreover, by the Hasse–Weilbound we know that |bi| �

√l. Thus

b1 ≡ ±b2 (mod p) and |b1 + b2|, |b1 − b2| � 2√

l < p

as l < p2/4. Thus, b1 = ±b2 and this completes the proof.

It remains to explain how this improves our computation. To apply Proposition 8.1 for somep we need to find a prime l satisfying conditions (a)–(c). The computationally expensive part isto compute al(E) = cl and al(ζ) for all ζ ∈ A(n, l). Let us, however, consider the applicationof Proposition 8.2 rather than Proposition 8.1. The computation proceeds as before by checkingconditions (a), (b) first. When we come to (c), we note that what we have to check is that{

�Eζ(Fl) �= l + 1 − al(E), if l ≡ 1 (mod 4),�Eζ(Fl) �= l + 1 ± al(E), if l ≡ 3 (mod 4),

for each ζ ∈ A(n, q). Rather than computing al(ζ) for each such ζ, we first pick a random pointin Eζ(Fl) and check whether it is annihilated by l + 1 − al(E) if p ≡ 1 (mod 4) and either of theintegers l + 1± al(E) if p ≡ 3 (mod 4). Only if this is the case do we need to compute al(ζ) to testcondition (c). In practice, for primes p ≈ 109, this brings a 10-fold speed-up in program run timefor Method II.

9. Eliminating exponents: Method III

Occasionally, Methods I and II fail to establish the non-existence of solutions to an equation of theform (10) for a particular exponent p even when it does seem that this equation has no solutions.The reasons for this failure are not clear to us. We, shall, however give a third method, rathersimilar in spirit to Kraus’ method (Method II), but requiring stronger global information furnishedby Proposition 3.1.

Suppose that D, d1, d2 are integers satisfying conditions (i)–(v) of Lemma 4.1. Let Et be one ofthe Frey curves associated with (10) and let f be a newform of the level predicted by Proposition 6.2

43



https://doi.org/10.1112/S0010437X05001739



with Fourier expansion as in (11), defined over a number field K. Define Tl(f) to be the set of τ ∈ Fl

such that either:

• p|NK/Q(al(Eτ ) − cl) and τ2 + d2 �≡ 0 (mod l); or• p|NK/Q(l + 1 ± cl) and τ2 + d2 ≡ 0 (mod l).

We suppose that −D is not a square and follow the notation of § 3. Fix a prime p satisfying (6).Suppose that l is a prime satisfying the following conditions.

(a) l �2D.(b) l = np + 1 for some integer n.(c) (−D2

l ) = 1, thus l splits in L = Q(√−D2), say (l) = l1l2.

(d) Each γ ∈ Γ is integral at l; what we mean by this is that each γ belongs to the intersection ofthe localizations Ol1 ∩ Ol2 .

We denote the two natural reduction maps by θ1, θ2 : Ol1 ∩ Ol2 → Fl. These of course correspondto the two squareroots for −D2 in Fl and are easy to compute.

Now let Γl be the set of γ ∈ Γ for which there exists τ ∈ Tl(f) such that:

• (d1τ + D1θ1(√−D2))n ≡ θ1(γ)n or 0 (mod l); and

• (d1τ + D1θ2(√−D2))n ≡ θ2(γ)n or 0 (mod l).

Proposition 9.1 (Method III). Let p be a prime satisfying condition (6). Let S be a set of primesl satisfying the conditions (a)–(d) above. With notation as above, if the newform f gives rise toa solution (t, s) to (10), then d1t + D1

√−D2 = γβp for some β ∈ O and some γ ∈

⋂l∈S Γl. In

particular, if⋂

l∈S Γl is empty, then the newform f does not give rise to any solution to (10) forthis exponent p.

Proof. Suppose that (t, s) is a solution to (10) arising from newform f via the Frey curve Et.Clearly θ1(t) = θ2(t) is simply the reduction of t modulo l. Let τ = θ1(t) = θ2(t) ∈ Fl. It followsfrom Lemma 6.3 that τ ∈ Tl(f). Let (x, y) be the solution to (9) corresponding to (t, s). Thusx = d1t. We know by Proposition 3.1 that

d1t + D1

√−D2 = γβp,

for some γ ∈ Γ and β ∈ O. Applying θi to both sides and taking nth powers (where we recall thatl = np + 1) we obtain

(d1τ + D1θi(√

−D2))n ≡ θi(γ)nθi(β)l−1 (mod l) with θi(β)l−1 ≡ 0 or 1 (mod l).

Thus γ ∈ Γl as defined above. The proposition follows.

10. Examples

It is clear that our three modular methods require computations of newforms of a given level. For-tunately the computer algebra suit MAGMA has a package completely devoted to such computations;the theory for these computations is explained by Cremona [Cre96] for rational newforms, and byStein [Ste05b] in the general case. As an alternative, we could have used Stein’s Modular FormsDatabase [Ste05a].

Example 2 (Absence of newforms). Lemma 4.1 and Proposition 6.2 lead us to associate solutions to(9), where p satisfies (6), with newforms of certain levels. If there are no newforms of the predictedlevels, we immediately deduce that there are no solutions to (9). With the help of a MAGMA programwe found all D = 1, 2, . . . , 100 where there are no newforms at the predicted levels. We deduce thefollowing result.

44



https://doi.org/10.1112/S0010437X05001739



Corollary 10.1. Let D be an integer belonging to the list

4, 16, 32, 36, 64.

Then (9) does not have any solutions with p satisfying condition (6).

This Corollary does not add anything new, as (1) has already been solved by Cohn’s methodfor D = 4, 16, 32, 36, 64 (but see [Ivo03, Sik03, Le02]).

Example 3. Corollary 5.2 solves (3) for all values of D in the range (2) except for 21 values; theseare the 19 values listed in (5) plus D = 55, 95. As indicated in § 2 the cases D = 55 and 95 havebeen solved by Bennett and Skinner. It is however helpful to look at the case D = 95 again asit shows how Methods I and III work together in harmony. There is only one possible signature(d1, d2) = (1, 95). Thus, t = x, s = y and we need to solve the equation

t2 + 95 = sp, where p � 7. (12)

As d1 = 1, it follows from Corollary 5.2 that y is even and so t = x is odd. Replacing t by −tif necessary, we can assume that t ≡ 1 (mod 4). Table 1 leads us to associate the solution (t, s, p)with the Frey curve

Et : Y 2 + XY = X3 +(

t − 14

)X2 +

(t2 + 95

64

)X.

From Proposition 6.2, we know that any solution to (12) arises from a newform of level 190. UsingMAGMA we find that there are, up to Galois conjugacy, precisely four newforms at level 190. Theseare

f1 = q − q2 − q3 + q4 − q5 + q6 − q7 + O(q8),

f2 = q + q2 − 3q3 + q4 − q5 − 3q6 − 5q7 + O(q8),

f3 = q + q2 + q3 + q4 + q5 + q6 − q7 + O(q8),

f4 = q − q2 + φq3 + q4 + q5 − φq6 + φq7 + O(q8), where φ2 + φ − 4 = 0.

The first three newforms above are rational and so correspond to the three isogeny classes ofelliptic curves of conductor 190. It turns out that none of these elliptic curves have non-trivial2-torsion. By the remarks made after Proposition 7.1 we know that Method I will be successful ineliminating all but finitely many exponents p. Indeed we find (in the notation of Proposition 7.1)that B3(f1) = B3(f3) = 15. Thus, we know that no solutions to (12) arise from the newformsf1 or f3, because otherwise, by Proposition 7.1, p|15 which contradicts p � 7. We also find thatB3(f4) = 24 × 3 and B7(f4) = 24 × 7. Thus no solution arises from f4. But,

B3(f2) = 3 × 7, B7(f2) = 32 × 5 × 7, B11(f2) = 0,

B13(f2) = 3 × 5 × 7 × 13, B17(f2) = 32 × 7 × 11.

We deduce that there are no solutions arising from f2 with exponent p > 7. It does, however, seemlikely that there is a solution with p = 7. Moreover, an attempt to prove that there is no solutionwith p = 7 using Method II fails: we did not find any integer 2 � n � 100 satisfying the conditionsof Proposition 8.1.

We apply Method III (and follow the notation of § 9). Write

ω =1 +

√−95

2.

Taking S = {113, 127, 239, 337, 491} we find that⋂l∈S

Γl ={−528 − 2ω

2187

}.

45



https://doi.org/10.1112/S0010437X05001739



Thus if we have any solutions at all then, by Proposition 9.1, we know that

(t − 1) + 2ω =(−528 − 2ω

2187

)(U + V ω)7,

for some integers U , V . Equating imaginary parts and simplifying we get

−U7 − 1855V U6 − 5061V 2U5 + 214 165V 3U4 + 416 605V 4U3

− 2 834 013V 5U2 − 2 944 375V 6U + 2818 247V 7 = 2187.

Using pari/gp we find that the only solution to this Thue equation is given by U = −3, V = 0.This shows that (t, s) = (529, 6).

The reader will notice that (t, s) = (−529, 6) is also a solution to (12) with p = 7 but it seemsto have been ‘missed’ by the method. This is not the case; we are assuming that the sign of t hasbeen chosen so that t ≡ 1 (mod 4). The solution (t, s) = (−529, 6) arises from some other newform(probably at some different level) and via a different Frey curve that we have not determined.

Example 4. For our last example we look at the case where D = 25. This, like 18 other cases, mustbe resolved by a combination of the modular approach and our lower bound for linear forms inthree logarithms which is to come. We assume that p � 7 and so p satisfies conditions (6). Thereare now two possible signatures (d1, d2) = (1, 25) and (5, 1) satisfying the conditions of Lemma 4.1.However, by Corollary 5.2, we may suppose that d1 > 1 and so d1 = 5, d2 = 1. We write t = x/5,s = y/5 where we know that t, s are integral by Lemma 4.1. Equation (10) becomes

t2 + 1 = 5p−2sp, t �= ±1.

Following Table 1, we associate with any solution to this equation the Frey curve

Et : Y 2 = X3 + 2tX2 − X,

and we know by Proposition 6.2 that any solution must arise from a newform of level 160. Using thecomputer algebra system MAGMA we find that there are, up to Galois conjugacy, three such newforms:

f1 = q − 2q3 − q5 − 2q7 + O(q8),

f2 = q + 2q3 − q5 + 2q7 + O(q8),

f3 = q + 2√

2q3 + q5 − 2√

2q7 + O(q8).

The first two newforms are rational, corresponding respectively to elliptic curves 160A1 and 160B1in Cremona’s tables [Cre96]. The third has coefficients in K = Q(

√2) and is straightforward to

eliminate using Method I. In the notation of Proposition 7.1 we find that if f3 does give rise to anysolutions (t, s, p) then p|B3(f3) = 24. This is impossible as p � 7 and so f3 does not give rise toany solutions.

We were unable to eliminate newforms f1 and f2 using Method I. Instead using our implemen-tation of Method II in MAGMA we showed that there are no solutions arising from either form with7 � p � 100. With our implementation of the improved Method II (Proposition 8.2) in pari/gpwe showed that there are no solutions with 100 � p � 163 762 845; this took roughly 26 hours on a2.4 GHz Pentium IV PC. The choice of where to stop the computation is of course not arbitrary, butcomes out of our bound for the linear form in logarithms. We will later prove that p � 163 762 845thereby completing the resolution of this case.

11. Results III

We applied the methods of the previous sections to solve all (3) with D is our range (2).

46



https://doi.org/10.1112/S0010437X05001739



Table 4. Computational details for Lemma 11.1 and its proof.

D (d1, d2) Ea p0 Machineb Time

7 (1, 7) 14A1 181 000 000 P1 26 h, 43 min15 (1, 15) 30A1 624 271 465 S1 252 h, 50 min18 (3, 2) 384D1, 384A1, 306 111 726 S3 293 h, 14 min

384G1, 384H123 (1, 23) 46A1 855 632 066 S2 477 h, 36 min25 (5, 1) 160A1, 160B1 163 762 845 P2 25 h, 58 min28 (2, 7) 14A1 315 277 186 P1 55 h, 41 min31 (1, 31) 62A1 860 111 230 S3 242 h, 2 min39 (1, 39) 78A1 852 830 725 P1 193 h, 41 min45 (3, 5) 480B1, 480F1, 340 749 424 S1 448 h, 43 min

480G1, 480H147 (1, 47) 94A1 1 555 437 629 S3 451 h, 34 min60 (2, 15) 30A1 358 541 296 S1 130 h, 30 min63 (1, 63) 42A1 292 825 735 S1 99 h, 45 min71 (1, 71) 142C1 2 343 468 548 S3 697 h, 26 min72 (3, 8) 96A1, 96B1 451 620 034 S1 316 h, 27 min79 (1, 79) 158E1 1 544 381 661 S3 448 h, 47 min87 (1, 87) 174D1 1 148 842 108 S3 329 h, 45 min92 (2, 23) 46A1 996 255 151 S3 285 h, 10 min99 (3, 11) 1056B1, 1056F1 593 734 622 P2 138 h, 46 min

100 (5, 4) 20A1 163 762 845 P1 21 h, 23 minaWe give here the Cremona code for the elliptic curves E as in his book [Cre96] andhis online tables: http://www.maths.nott.ac.uk/personal/jec/ftp/data/INDEX.html.bThe machines are as follows: P1, 2.2 GHz Intel Pentium PC; P2, 2.4 GHz Intel PentiumPC; S1, Dual processor 750MHz UltraSPARC III; S2, 650 MHz UltraSPARC IIe; S3,UltraSPARCIII with 12 processors of 1050 MHz speed.

Lemma 11.1. Suppose that 1 � D � 100 and p is a prime satisfying (6). If (x, y, p) is a solutionto (3) that is not included in the table in Appendix A, then D is one of

7, 15, 18, 23, 25, 28, 31, 39, 45, 47, 60, 63, 71, 72, 79, 87, 92, 99, 100. (13)

Moreover, (x, y, p) has signature (d1, d2) and arises from an elliptic curve E and p > p0 where E,p0 and (d1, d2) are given by Table 4.

Proof. We wrote a MAGMA program that does the following. For each D in the range (2) we writedown the set of possible signatures (d1, d2) satisfying the conditions of Lemma 4.1.

For each such pair (d1, d2) write down the (one or two) Frey curves given by Tables 1–3, bearingin mind the information given by Corollary 5.2.

For each Frey curve we compute the conductor (given by Proposition 6.2) of the newforms givingrise to possible solutions and then write down all of these newforms.

We attempt to eliminate each newform f using Method I. This involves searching for primesl �2D such that (in the notation of Proposition 7.1) Bl(f) �= 0. If we are successful and find suchprimes l1, . . . , lm, then by Proposition 7.1 this exponent divides all of the Bli(f), and so dividestheir greatest common divisor B (say). If B is divisible by any prime p that satisfies condition (6),then we attempt to eliminate this possible p using Method II: this involves searching for an integer2 � n � 100 satisfying conditions (a)–(c) of Proposition 8.1. If one such n is found then we knowthat there are no solutions for the exponent p. Otherwise, we apply Method III (Proposition 9.1)to write down Thue equations leading to possible solutions.

47



https://doi.org/10.1112/S0010437X05001739



As predicted by the comments made after Proposition 7.1, Method I succeeded with all non-rational newforms and all rational newforms corresponding to elliptic curves with only trivial2-torsion (it also succeeded with some rational newforms corresponding to elliptic curves withnon-trivial 2-torsion). Indeed, we found no solutions arising from non-rational newforms for D inour range 1 � D � 100.

We are left only with rational newforms f that correspond to elliptic curves E having somenon-trivial 2-torsion. The details of these are documented in Table 4. For primes p < 100 satisfyingcondition (6) we attempt to show that there are no solutions arising from E for the particularexponent p using Method II (as before). If this fails for a particular exponent p, then we useMethod III to write down the Thue equations leading to the possible solutions.

Our proof that p � 100 is now complete except that there are some Thue equations to solve. Wehad to solve Thue equations of degree 7 for D = 7, 47, 79 and 95. These were solved using pari/gpand the solutions are incorporated in the table in Appendix A. We also had to solve a Thue equationof degree 11 for D = 23, of degree 17 for D = 28 and of degree 13 for D = 92. We were unableto (unconditionally) solve these three Thue equations using the built-in functions of pari/gp. Thereason is that, in each case, it was impossible for pari/gp to prove that the system of units it hadfound, although of correct rank, was maximal. We are grateful to Dr. Guillaume Hanrot for sendingus his pari program for solving Thue equations without the full unit group. This program, basedon [Han00], solved all three equations in a few minutes.

For the next step we implemented our improved Method II (Proposition 8.2) in pari/gp (seethe remark after the proof). To complete the task and show that p > p0 for any missing solutionwe used our pari/gp program to disprove the existence of any missing solution for each prime100 � p � p0. We ran this pari/gp program on various machines as indicated in Table 4. The totalcomputer time for this step is roughly 206 days.

Remark. The reader may be surprised that some of the computations were done in MAGMA whereasothers were carried out in pari/gp. As stated earlier, MAGMA has a package for computing modularforms. This is essential for us and is unavailable in pari/gp.

For showing that p > p0, it is simply not practical to use MAGMA. Here we are using the improvedMethod II (Proposition 8.2). The main bottleneck in Method II is computing al(E) for primes lthat can be about 1011 (recall l is a prime satisfying l ≡ 1 (mod p)). For this pari/gp uses thetheoretically slower Shanks–Mestre method [Coh93a] rather than the theoretically faster Schoof–Elkies–Atkin [Sch95b] method used by MAGMA. However, for primes of the indicated size it seemsthat pari/gp is about 10 times faster than MAGMA.

The reader may also note that two of the machines we used are multiprocessor machines. Thecomputation for each D could have been speeded up considerably by parallelising. We howeverdecided against this, so as to keep our programs simple and transparent.

12. The ‘modular’ lower bound for y

In this section we would like to use the modular approach to prove a lower bound for y with D inthe range (2). Before doing this we prove a general result for arbitrary non-zero D.

Proposition 12.1. Suppose that D is a non-zero integer, and d1, d2 satisfy Lemma 4.1(i)–(v).Suppose that (t, s, p) is a solution to (10) arising from a rational newform f via a Frey curve Et.Then either rad(s)|2d1 or |s| > (

√p − 1)2.

Proof. As the newform is rational we know that the newform f corresponds to an elliptic curveE/Q whose conductor equals the level of f .

48



https://doi.org/10.1112/S0010437X05001739



Suppose that rad(s) does not divide 2d1. As t and d2 are coprime we see that there is some primel|s so that l �2D. By Lemma 6.4 we see that p divides l + 1± al(E). It follows from the Hasse–Weilbound that l + 1 ± al(E) �= 0, and so

p � l + 1 ± al(E) < (√

l + 1)2,

again using Hasse–Weil. Thus l > (√

p − 1)2. The proposition follows as l|s.Corollary 12.2. Suppose that D is one of the values in (13). If (x, y, p) is a solution to (9) not inthe table in Appendix A, then y > (

√p − 1)2.

Proof. Suppose that D is in the range (2) and (x, y, p) is some solution to (9) not in the table inAppendix A. From the preceding sections we know that this solution must satisfy condition (6).Moreover by Lemma 4.1, x = d1t and y = rad(d1)s, where (t, s, p) satisfy (10) for some d1 and d2

satisfying conditions (i)–(v) of that lemma.We have determined, for 1 � D � 100, all solutions to (10) arising from non-rational newforms

(indeed there were none). Thus we may suppose that our putative solution arises from a rationalnewform. By Proposition 12.1 we see that either |y| � |s| > (

√p− 1)2 or rad(s)|2d1. We must prove

that the second possibility does not arise.Suppose that rad(s)|2d1. From Lemma 4.1 we see that rad(y)|2d1. We first show that rad(y) �= 2.

For in this case we have reduced to an equation of the form x2 + D = 2m. For |D| < 296, Beukers[Beu81, Corollary 2] shows that m � 18 + 2 log |D|/ log 2. A short MAGMA program leads us to allthe solutions to this equation for 1 � D � 100 and we find that these are already in our table inAppendix A.

Thus, we may suppose that rad(y)|2d1 and rad(y) �= 2. An examination of the possible casesreveals the following possibilities

D = 18, 45, 72, 99 and rad(y) = 3, D = 25, 100 and rad(y) = 5.

On removing the common factors, each case quickly reduces to an equation that has already beensolved. For example, we must solve x2 + 100 = yp under the assumption that rad(y) = 5 orequivalently the equation x2 + 100 = 5m. Removing the common factor reduces to the equationX2 + 4 = 5m−2. However, X2 + 4 = Y n has already been solved and only has the solutions(X,Y, n) = (2, 2, 3), (11, 5, 3). We quickly see that the only solution to x2 + 100 = 5p is (x, p) =(55, 5).

13. The linear form in logarithms

It is useful at this point to recap what we have done so far. We would like to complete the proof ofTheorem 1 by showing that our table in Appendix A is not missing any solutions. So let us supposethat our table in Appendix A is missing some solution (x, y, p) to (3) for some value of D in ourrange (2). We have proved (Lemma 11.1) that D is one of the values in (13). Moreover (again byLemma 11.1 and by Corollary 12.2), any missing solution (x, y, p) must satisfy

p > p0, y � (√

p − 1)2, (14)

with p0 given by Table 4. Our aim is to show that p � p0: a contradiction.From the table of values of p0 we know that

|x|, p � 108 (15)

and indeed much more, although this inequality is sufficient for much of our later work. In theremainder of this paper we assume that D is one of the remaining values (13), and always write

49



https://doi.org/10.1112/S0010437X05001739



(as before) D = D21D2, where D2 is square-free. The triple (x, y, p) will always be a solution to (3)

supposedly missing from our table in Appendix A and hence satisfying the above inequalities.In this section we write down the linear form in logarithms corresponding to (3) and apply

a theorem of Matveev to obtain upper bounds for the exponent p. These upper bounds obtainedfrom Matveev’s theorem are not small enough to contradict our lower bounds for p obtained inLemma 11.1, but they are needed when we come to apply our bounds for linear forms in threelogarithms given in the next section.

Lemma 13.1. Let (d1, d2) be the signature of our supposedly missing solution (x, y, p) (which weknow from Lemma 11.1). Define

d =

{d1, for D �≡ 7 (mod 8),2d1, for D ≡ 7 (mod 8).

(16)

Then d is a prime power, say d = qc for some prime q, where, moreover, q splits in L = Q(√−D2), say

(q) = qq. Let k0 be the smallest positive integer such that the ideal qk0 is principal, say qk0 = (α0).Also let

k =

{k0, if k0 is odd,

k0/2, if k0 is even,and κ =

{2, if k0 is odd,

1, if k0 is even,so that k =

κk0

2.

Then there exists γ ∈ L such that(x − D1

√−D2

x + D1

√−D2

)k

= ακγp, where α = α0/α0, h(α) =k0 log d

2, h(γ) =

k log y

2.

Proof. We begin with the factorization

(x + D1

√−D2)(x − D1

√−D2) = yp.

Our first step is to show that any prime divisor q of y splits in L. Suppose otherwise, then we maywrite (q) = q or (q) = q2 for some prime ideal q satisfying q = q. If p = 2r+1 then clearly qr dividesboth factors on the left-hand side above and so divides 2D1

√−D2. This is impossible in view of

the fact that p is enormous and 1 � D � 100. Thus, we have shown that every prime divisor q of ysplits in L. Put

y =∏i∈I

qiai and (qi) = qiqi, qi �= qi, i ∈ I, then (x + D1

√−D2) =

∏i∈I

(qibi qci

i ),

where we assume (for ease of notation) that bi � ci for all i. Thus

(x − D1

√−D2) =

∏i∈I

(qici qbi

i ), with bi + ci = pai, for all i ∈ I.

Then, clearly,

d := gcd(x + D1

√−D2, x − D1

√−D2) =

∏i∈I

(qiqi)ci =∏i∈I

(qi)ci .

This shows that d = (d) where d ∈ Z. We would like to calculate this d and verify that its valueis in agreement with (16). From the definition of d we see that d|2x and d|2D1. However, by ourdefinition of signature, gcd(x2,D) = d2

1. It follows that d2|4d21 and so d|2d1. However, d1|x and

d1|D1. Hence, d1|d and so d1|d. Thus, d = d1 or d = 2d1. We note the following cases.

• If D2 �≡ 7 (mod 8) then 2 �y. Thus 2 �d and so d = d1.

• Suppose that D2 ≡ 7 (mod 8). Now from Lemma 4.1 and its proof we know that D = d21d2

and x = d1t where gcd(t, d2) = gcd(d1, d2) = 1. Clearly d2 = d23D2 with d3 = D1/d1 integral.

50



https://doi.org/10.1112/S0010437X05001739



Suppose first that d1 is even. It follows easily that t, d2 are odd and

(d) = d = 2d1

(t + d3

√−D2

2,t − d3

√−D2

2

).

Hence (2d1)|d and so d = 2d1.• The only case left to consider is D2 ≡ 7 (mod 8) and d1 is odd. By examining Table 4 we see

that d1 = 1. Thus, 2|y by Corollary 5.2. Clearly x is odd and the same argument as aboveshows that d = 2 = 2d1.

This proves that d satisfies (16). By looking again at the possible values of d1 in Table 4 we see thatd is a prime power in all cases. Let j ∈ I such that d = q

cj

j . Thus, ci = 0 for all j �= i. Then

(x + D1

√−D2) = q

cj

j · qjbj ·

∏j �=i

qipai ,

whence(x + D1

√−D2) = (qjqj

−1)cj ·∏i∈I

qipai = (aa−1)gp,

where a and g are integral ideals with a = qcj

j , N (a) = qjcj = d, N (g) = y, and N denotes the

norm. Thus, as ideals, (x − D1

√−D2

x + D1

√−D2

)= (aa−1)2(gg−1)p.

We define k0, k, κ, α0 as in the statement of the lemma. Thus, ak0 = (α0) and we have therelation (between ideals)

(x + D1

√−D2)k = (a/a)kgkp = a2k(N (a))−kgkp = (α0)κ(d)−kgkp.

However, p is an enormous prime certainly not dividing the class number. This shows that gk isalso principal, gk = (γ0) say, where γ0 is an algebraic integer chosen so that the following equalityof elements of L holds

(x + D1

√−D2)k = ακ

0d−kγp0 , with N (α0) = dk0 and N (γ0) = yk.

Put α = α0/α0 and γ = ±γ0/γ0. The proof of the lemma is complete except for the statementsabout the heights of α, γ. These follow from Lemma 13.2 below.

Lemma 13.2. Let α be an algebraic number whose conjugates are all (including α itself) of modulusequal to 1, then h(α) = (log a)/deg α, where a is the leading coefficient of the minimal polynomialof α. In particular, if α = α0/α0 where α0 is a non-real quadratic irrationality, then h(α) =12 logN (α0).

Proof. Set d = deg α. By hypothesis α is a root of a polynomial of Z[X] of the form P (X) =aXd + · · · . We have h(α) = 1

d log M(P ), where M is Mahler’s measure, and the first result easilyfollows because the roots of P are of modulus 1. This proves the first assertion. The (easy) proof ofthe second assertion is omitted.

We now write the linear form in three logarithms. Define

Λ = log(

x − D1

√−D2

x + D1

√−D2

),

where we have taken the principal determination of the logarithm.

Lemma 13.3. We have

log |Λ| � −p

2log y + log(2.2D1

√D2).

51



https://doi.org/10.1112/S0010437X05001739



Proof. We will rely on the lower bounds (15). Clearly∣∣∣∣x − D1

√−D2

x + D1

√−D2

− 1∣∣∣∣ < 2

D1

√D2

|x| .

A standard inequality [Sma98, Lemma B.2] shows that

|Λ| < 2.1D1

√D2

|x| ,

so that

log |Λ| < −log|x| + log(2.1D1

√|D2|).

Using the fact that yp − x2 = D and a similar argument to that above, we deduce the lemma.

To bound p we use the theory of linear forms of (at most three) logarithms. We need the specialcase of three logarithms of a theorem of Matveev.

Theorem 2 (Matveev). Let λ1, λ2, λ3 be Q-linearly independent logarithms of non-zero algebraicnumbers and let b1, b2, b3 be rational integers with b1 �= 0. Define αj = exp(λj) for j = 1, 2, 3 and

Λ = b1λ1 + b2λ2 + b3λ3.

Let D be the degree of the field Q(α1, α2, α3) over Q. Put χ = [R(α1, α2, α3) : R]. Let A1, A2, A3

be positive real numbers, which satisfy

Aj � max{Dh(αj), |λj |, 0.16} (1 � j � 3).

Assume that B � max{1,max{|bj |Aj/A1; 1 � j � 3}}. Then

log |Λ| > −C1D2A1A2A3 log(1.5eDB log(eD)),

where

C1 =5 × 165

6χe3(7 + 2χ)

(3e2

)χ

(20.2 + log(35.5D2 log(eD))).

In particular, for D = 2 and χ = 2, this gives

log |Λ| > −1.807 41 × 1011A1A2A3 log(13.807 36B). (17)

For a proof see [Mat00].

13.1 A preliminary bound for p

It follows from Lemma 13.1 that

kΛ = κ log α + p log γ + iqπ = κ log α + p log γ + r log(−1), r ∈ Z,

which appears as a linear form of logarithms. However, a small transformation of this form leads tobetter estimates. Write

kΛ = κ log(ε1α) + p log(ε2γ) + iqπ, q ∈ Z,

where ε1 and ε2 = ±1 are chosen so that |log(ε1α)| < π/2 and |log(ε2γ)| < π/2, where we takeprincipal values for the logarithms and q such that |Λ| is minimal.

Remark. Indeed, we can take any roots of unity in L for ε1 and ε2. The only relevant case forour set of outstanding values of D are D = 25, 100, where L = Q(

√−1), whence we can realize

|log(ε1α)| < π/4 and |log(ε2γ)| < π/4, and we write

Λ = 2 log α + p log γ + q log ζ, where ζ = eiπ/2.

52



https://doi.org/10.1112/S0010437X05001739



We now return to the general case. By Lemma 13.3

log |kΛ| � −p

2log y + log(2.2kD1

√D2).

Our lower bounds for x, y and p imply that log |kΛ| is very small and it is straightforward to deducethat |r| � (p + 1)/2. We can write kΛ in the form

kΛ = b1λ1 + b2λ2 + b3λ3

with b1 = κ (= 1 or 2), α1 = ε1α, b2 = p, α2 = ε2γ, b3 = q, α3 = −1 and

h(α1) =k

κlog d, λ1 = log α1, h(α2) =

k log y

2, |λ2| < π/2, h(α3) = 0, λ3 = iπ,

except for the case L = Q(√−1) studied in the previous remark where λ3 = iπ/2.

Applying Theorem 2, we have D = χ = 2 and we can take

A1 = max{

2k log d

κ,π

2

}, A2 = max

{k log y,

π

2

}, A3 = π

and (using some change of notation in Theorem 2) B = p + 1 (this choice of B is justified by theinequality |q| � (p + 1)/2 proved above), and we get

p � C2k2 log(2D1) log p.

This implies p � C3k2 log(2D1) log(k2 log(2D1)), and thus

p � C4D2 log(2D1) log(D2 log(2D1)), (18)

where the constants are easily made explicit.

Lemma 13.4. Suppose that D is one of the remaining values (13) and (x, y, p) is a solution to (9)missing from our table in Appendix A.

• If D = 7 then p < 6.81 × 1012.

• Otherwise, if D is square-free, then p < 1.448 × 1015.

• For other values of D, we have p < 3.966 × 1014.

Proof. This is a simple application of Matveev’s Theorem 2. If D = 7 it is easy to show that the α0

arising in Lemma 13.1 is (up to conjugation) (1 +√−7)/2, we know that k = 1; thus N (α0) = 2

and (log α0) = 1.209 429 202 8 . . . . Then we can apply (17) with A1 = π/2, A2 = log y, log A3 = πand B = p + 1. After a few iterations we get the stated bound on p.

In the application of Theorem 2, we can take, for all the square-free values of D,

A1 =

{7 log 2, if k0 is odd,8 log 2, if k0 is even,

A2 =

{7 log y, if k0 is odd,4 log y, if k0 is even,

A3 = π,

so that A1A2 � 49 log 2 × log y and then we get p < 1.448 × 1015.

For all the remaining values of D, we can take

A1 =

log 10, if h = 1,π/2, if h = 2,3 log 2, if h = 3,

A2 =

log y, if h = 1,log y, if h = 2,3 log y, if h = 3,

A3 = π,

so that A1A2 � 9 log 2 × log y, and we get now p < 3.966 × 1014.

53



https://doi.org/10.1112/S0010437X05001739



14. A new estimate on linear forms in three logarithms

14.1 Statement of the resultWe shall apply the following theorem.

Theorem 3. We consider three non-zero algebraic numbers α1, α2 and α3, all �= 1, which are eitherall real or all complex of modulus one. Moreover, we assume that{

either α1, α2 and α3 are multiplicatively independent, or

two multiplicatively independent, the third a root of unity �= 1.(M)

We also consider three non-zero rational integers b1, b2, b3 with gcd(b1, b2, b3) = 1 and the linearform

Λ = b1 log α1 + b2 log α2 + b3 log α3 �= 0,where the log αi are arbitrary determinations of the logarithm, but which are all real or all purelyimaginary. Without loss of generality, we assume that

b2|log α2| = |b1 log α1| + |b3 log α3| ± |Λ|.Let K, L, R, R1, R2, R3, S, S1, S2, S3, T , T1, T2, T3 be rational integers that are all � 3, with

L � 5, R > R1 + R2 + R3, S > S1 + S2 + S3, T > T1 + T2 + T3.

Let ρ > 2 be a real number. Assume first that(KL

2+

L

4−1− 2K

3L

)log ρ � (D+1) log N +gL(a1R+a2S+a3T )+D(K−1) log b−2 log(e/2), (19)

where N = K2L, D = [Q(α1, α2, α3) : Q]/[R(α1, α2, α3) : R], e = exp(1),

g =14− N

12RST, b = (b2η0)(b2ζ0)

(K−1∏k=1

k!)−4/K(K−1)

,

with

η0 =R − 1

2+

(S − 1)b1

2b2, ζ0 =

T − 12

+(S − 1)b3

2b2,

and

ai � ρ|log αi| − log |αi| + 2Dh(αi), i = 1, 2, 3.If, for some positive real number χ and V := ((R1 + 1)(S1 + 1)(T1 + 1))1/2:

(i) (R1 + 1)(S1 + 1)(T1 + 1) > K max{R1 + S1 + 1, S1 + T1 + 1, R1 + T1 + 1, χV };(ii) Card{αr

1αs2α

t3 : 0 � r � R1, 0 � s � S1, 0 � t � T1} > L;

(iii) (R2 + 1)(S2 + 1)(T2 + 1) > 2K2;(iv) Card{αr

1αs2α

t3 : 0 � r � R2, 0 � s � S2, 0 � t � T2} > 2KL; and

(v) (R3 + 1)(S3 + 1)(T3 + 1) > 6K2L;

then either

Λ′ > ρ−KL,

where

Λ′ = |Λ| · max{

LReLR|Λ|/(2b1)

2|b1|,LSeLS|Λ|/(2b2)

2|b2|,LTeLT |Λ|/(2b3)

2|b3|

},

or at least one of the following conditions (C1), (C2), (C3) hold:

|b1| � Ri and |b2| � Si and |b3| � Ti, (i = 1, 2) (Ci)

54



https://doi.org/10.1112/S0010437X05001739



(C3) either there exist two non-zero rational integers r0 and s0 such that

r0b2 = s0b1,

with

|r0| � BS :=(R1 + 1)(T1 + 1)

χV − max{R1, T1}and |s0| � BR :=

(S1 + 1)(T1 + 1)χV − max{S1, T1}

,

or there exist rational integers r1, s1, t1 and t2, with r1s1 �= 0, such that

(t1b1 + r1b3)s1 = r1b2t2, gcd(r1, t1) = gcd(s1, t2) = 1,

which also satisfy, for δ = gcd(r1, s1),

0 < |r1s1|/δ � BT :=(R1 + 1)(S1 + 1)

χV − max{R1, S1}, |s1t1|/δ � BR and |r1t2|/δ � BS.

Proof. A detailed proof can be found in [Mig]. It contains many technical improvements whencompared with the result proved in [BMS], but the main progress is a zero-lemma due to Laurent[Lau03], which improves [Gou02] and provides an important improvement on the zero-lemma usedin our previous paper [BMS].

14.2 How to use Theorem 3To apply the theorem, we consider an integer L � 5 and real parameters m > 0, ρ > 2 (then onecan define the ai) and we put

K = �mLa1a2a3�, with ma1a2a3 � 2.

To simplify the presentation, we also assume m � 1 and a1, a2, a3 � 1, and put

R1 = �c1a2a3�, S1 = �c1a1a3�, T1 = �c1a1a2�,R2 = �c2a2a3�, S2 = �c2a1a3�, T2 = �c2a1a2�,R3 = �c3a2a3�, S3 = �c3a1a3�, T3 = �c3a1a2�,

where the ci satisfy the conditions (i)–(v) of the theorem.Clearly, condition (i) is satisfied if

(c31(a1a2a3)2)1/2 � χma1a2a3L, c2

1 · a � 2mL, where a = min{a1, a2, a3}.Condition (ii) is true when 2c2

1a1a2a3 · min{a1, a2, a3} � L; we can take

c1 = max{(χmL)2/3, (2mL/a)1/2}.To satisfy (iii) and (iv) we can take

c2 = max{21/3(mL)2/3,√

m/aL}.Finally, because of the hypothesis (M), condition (v) holds for

c3 = (6m2)1/3L.

Remark. When α1, α2, α3 are multiplicatively independent then it is enough to take c1 and c3 asabove and c2 = 21/3(mL)2/3.

Then we have to verify the condition (19). When this inequality holds, one obtains the lowerbound |Λ′| > ρ−KL and we get

log |Λ| > −KL log ρ − log(max{R,S, T} · L),

except maybe if at least one of the conditions (C1), (C2) or (C3) holds.

55



https://doi.org/10.1112/S0010437X05001739



15. Completion of the proof of Theorem 1

Having given our new bounds for linear forms in three logarithms we now use them to completethe proof of Theorem 1. We have indeed shown in Lemma 11.1 that if (x, y, p) is a missing solutionthen p > p0 where p0 is given in Table 4. To complete the proof it is enough to show that p � p0.In § 13 we wrote down the linear form in logarithms we obtain for each outstanding value of D.We will content ourselves by giving the details of this calculation for D = 7. The other cases arepractically identical (but with different constants).

We have seen in Lemma 13.3 that

Λ := logx −

√−7

x +√−7

satisfies log |Λ| � −p

2log y + log(2.2

√7).

Writing α0 = (1 +√−7)/2 we saw that the linear form is given by

Λ = 2 log(ε1α0/α0) + p log(ε2γ/γ) + iqπ, ε1, ε2 = ±1,

for some rational integer q with |q| < p and we get

log |Λ| > −KL log ρ − log(max{R,S, T} · L),

except maybe if at least one of the conditions (C1), (C2) or (C3) holds.Now we proceed effectively to the computation of an upper bound for p. The first step is to

recall that we have proved in Lemma 13.4, by applying Matveev’s theorem (Theorem 2), that

p < 6.81 × 1012.

We then apply our Theorem 3 with the initial condition p < 6.81 × 1012 and with the lowerbound y � 22; note that we do not yet assume our lower bound (14) obtained through the modularapproach. There are two reasons for this.

• The first reason is that we would like to demonstrate how powerful our new lower bound forlinear forms in three logarithms is, even without the help of the modular approach.

• The second reason is that when we later make the assumption (14), and apply our lowerbound for linear forms in three logarithms, the reader will appreciate the saving brought bythe ‘modular lower bound’ for y.

So for now we assume simply that y � 22, which holds because y is even, not a power of 2 and that−7 is a quadratic residue for every odd prime factor of y (see [Les98]). In a few steps we can provethat

p < 4.2 × 108.

The reader should compare this bound with the bound p < 6.81 × 1012 obtained by Matveev’stheorem.

We now assume our ‘modular’ lower bound for y in (14), with p > 1.3 × 108, and then we shallobtain a much better bound for p. We give much more detail.

We have to distinguish two cases

b1 = 2, α1 = ε′α/α, b2 = p, α2 = εγ/γ, b3 = q, α3 = −1, (I)

and

b1 = 2, α1 = ε′α/α, b2 = q, α2 = −1, b3 = p, α3 = εγ/γ. (II)

Let us first consider case (I). Applying Theorem 3 we get

p < 4.3 × 108

56



https://doi.org/10.1112/S0010437X05001739



with the choices L = 110, ρ = 6, m = 71.602 265 32, χ = 0.7 and

R1 = 178 896, S1 = 29587, T1 = 47734, R2 = 285 899, S2 = 47284, T2 = 76285and R3 = 1975 684, S3 = 326 756, T3 = 527 164,

unless at least one of the conditions (C1), (C2), (C3) holds. For these values, it is clear that, becausesince we know that p > 108, conditions (C1) and (C2) do not hold.1 The values of BR and BT definedin Theorem 3 are equal to

BR = 127, BT = 483.

The first case of condition (C3) implies that p � BR = 127, a contradiction. Thus, we have toconsider the last alternative:

(t1b1 + r1b3)s1 = r1t2b2.

Putting r1 = δr′1 and s1 = δs′1 and simplifying, we get here

(2t1 + δr′1q)s′1 = r′1t2p.

which shows that s′1 = 1 and r′1 = 1 or 2, and gives

(2/r′1)t1 + δq = t2p, with |t2| � BT /2.

We write δΛ = δb1 log α1 + δb2 log α2 + (t2b2 − (b1/r′1)t1) log α3, that is

δΛ =2r′1

log(αr′1δ1 /αt1

3 ) + p log(αδ2α

t23 ) = a linear form in two logarithms

and we apply [LM95] (with L = 10 and ρ = 16), which now gives p < 3 × 108.Thus, we have proved that, in case I,

p < 4.3 × 108.

Concerning case (II), we first note that in the non-degenerate case we obtain p < 4.3 × 108 asbefore. Then we note that (C1) or (C2) implies that p � max{T1, T2}, (the present Ti play the roleof the previous Si, and both are bounded independently of y.) Now we study condition (C3). Forthe first alternative r0b2 = s0b1, we get |q| < BR and we can apply [LM95] to the linear form in twologarithms

Λ = (log(ε′α/α)2 + q log(−1)) + p log(εγ/γ),

which works quite well. Consider now the second alternative, which gives here (we have t2 �= 0: ift2 = 0, then p < 108)

(2/r′1)t1 + δp = t2q′, with |s1| � BS/2 and q = s′1q

′, r′1 = 1 or 2.

We write now t2Λ = t2b1 log α1 + t2b2 log α2 + (s1b3 + (b1/r′1)s

′1t1) log α3, that is

t2Λ =2r′1

log(αr′1t21 α

s′1t13 ) + p log(αt2

2 αs13 )

and we apply [LM95] (again with L = 10 and ρ = 16), which now gives p < 2 × 108.Thus, we have proved that, in all cases,

p < 4.3 × 108.

1To be more precise we can take the above values for S1, S2 and S3 independently of y but the Ri and Ti have to beincreased for y > 1.3 × 108, as can be seen on the definition of the parameters given in the previous section (a1 anda3 are independent of y but not a2). Luckily, the larger y is, the better our resulting estimate for p will be and thuswe can always replace y by some lower bound for it.

57



https://doi.org/10.1112/S0010437X05001739



Iterating this process four times we obtain that, in all cases

p < 1.3 × 108,

which is indeed better than the upper bound used in the modular computation.

Remark. We note that without the modular lower bound for y we were able to show that p <1.11 × 109, but with this modular lower bound we were able to improve this to p < 1.81 × 108.Whilst it is certainly possible to reach the former target with the methods of this paper, it wouldhave taken about six times as long as it took to reach the latter. From this it is a plausible guessthat without the modular lower bound for y the computational part for the entire proof for all thevalues of 1 � D � 100 might have taken at least 800 days rather than 206 days.

Acknowledgements

We would like to warmly thank Mihai Cipu and Attila Petho for pointing many imperfections ina previous version of this paper, and Guillaume Hanrot for help with solving Thue equations. Weare also grateful to the anonymous referee for many pertinent historical remarks and a very carefulreading.

Appendix A

We list all the solutions to (1) in the range 1 � D � 100.

D Solutions (|x|, |y|, n)

1 (0, 1, n)2 (5, 3, 3)34 (2, 2, 3), (11, 5, 3)567 (1, 2, 3), (181, 32, 3), (3, 2, 4), (5, 2, 5), (181, 8, 5), (11, 2, 7), (181, 2, 15)8 (0, 2, 3)9

1011 (4, 3, 3), (58, 15, 3)12 (2, 2, 4)13 (70, 17, 3)1415 (7, 4, 3), (1, 2, 4), (7, 2, 6)16 (0, 2, 4), (4, 2, 5)17 (8, 3, 4)18 (3, 3, 3), (15, 3, 5)19 (18, 7, 3), (22434, 55, 5)20 (14, 6, 3)212223 (2, 3, 3), (3, 2, 5), (45, 2, 11)2425 (10, 5, 3)26 (1, 3, 3), (207, 35, 3)27 (0, 3, 3)28 (6, 4, 3), (22, 8, 3), (225, 37, 3), (2, 2, 5), (6, 2, 6), (10, 2, 7), (22, 2, 9), (362, 2, 17)2930

58



https://doi.org/10.1112/S0010437X05001739




31 (15, 4, 4), (1, 2, 5), (15, 2, 8)32 (7, 3, 4), (0, 2, 5), (88, 6, 5)333435 (36, 11, 3)36373839 (5, 4, 3), (31, 10, 3), (103, 22, 3), (5, 2, 6)40 (52, 14, 3)41424344 (9, 5, 3)45 (96, 21, 3), (6, 3, 4)4647 (13, 6, 3), (41, 12, 3), (500, 63, 3), (14, 3, 5), (9, 2, 7)48 (4, 4, 3), (148, 28, 3), (4, 2, 6)49 (524, 65, 3), (24, 5, 4)50515253 (26, 9, 3), (156, 29, 3), (26, 3, 6)54 (17, 7, 3)55 (3, 4, 3), (419, 56, 3), (3, 2, 6)56 (76, 18, 3), (5, 3, 4)57585960 (2, 4, 3), (1586, 136, 3), (14, 4, 4), (50 354, 76, 5), (2, 2, 6), (14, 2, 8)61 (8, 5, 3)6263 (1, 4, 3), (13 537, 568, 3), (31, 4, 5), (1, 2, 6), (31, 2, 10)64 (0, 4, 3), (0, 2, 6), (8, 2, 7)65 (4, 3, 4)6667 (110, 23, 3)68697071 (21, 8, 3), (35, 6, 4), (46, 3, 7), (21, 2, 9)72 (12, 6, 3), (3, 3, 4)7374 (985, 99, 3), (13, 3, 5)7576 (7, 5, 3), (1015, 101, 3)77 (2, 3, 4)7879 (89, 20, 3), (7, 2, 7)80 (1, 3, 4)81 (46, 13, 3), (0, 3, 4)8283 (140, 27, 3), (140, 3, 9)

59



https://doi.org/10.1112/S0010437X05001739




84858687 (16, 7, 3), (13, 4, 4), (13, 2, 8)8889 (6, 5, 3)909192 (6, 2, 7), (90, 2, 13)939495 (11, 6, 3), (529, 6, 7)96 (23, 5, 4)97 (48, 7, 4)9899 (12, 3, 5)

100 (5, 5, 3), (30, 10, 3), (198, 34, 3), (55, 5, 5)

References

Ape60a R. Apery, Sur une equation diophantienne, C. R. Acad. Sci. Paris Ser. I Math. 251 (1960),1263–1264.

Ape60b R. Apery, Sur une equation diophantienne, C. R. Acad. Sci. Paris Ser. I Math. 251 (1960),1451–1452.

BBBCO C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier, User’s guide to PARI-GP, version2.1.1, http://pari.math.u-bordeaux.fr/.

Ben04 M. A. Bennett, Products of consecutive integers, Bull. London Math. Soc. 36 (2004), 683–694.BS04 M. A. Bennett and C. M. Skinner, Ternary Diophantine equations via Galois representations and

modular forms, Canad. J. Math. 56 (2004), 23–54.Beu81 F. Beukers, On the generalized Ramanujan–Nagell equation I, Acta Arith. 38 (1981), 389–410.BH96 Yu. Bilu and G. Hanrot, Solving Thue equations of high degree, J. Number Theory 60 (1996),

373–392.BHV01 Yu. Bilu, G. Hanrot and P. M. Voutier, Existence of primitive divisors of Lucas and Lehmer

numbers. With an appendix by M. Mignotte. J. reine angew. Math. 539 (2001), 75–122.BCP97 W. Bosma, J. Cannon and C. Playoust, The Magma Algebra System I: The User Language,

J. Symbolic Comput. 24 (1997), 235–265, http://www.magma.maths.usyd.edu.au/magma/.BCDT01 C. Breuil, B. Conrad, F. Diamond and R. Taylor, On the modularity of elliptic curves over Q:

wild 3-adic exercises, J. Amer. Math. Soc. 14 (2001), 843–939.BL96 Y. Bugeaud and M. Laurent, Minoration effective de la distance p-adique entre puissances de

nombres algebriques, J. Number Theory 61 (1996), 311–342.BMS Y. Bugeaud, M. Mignotte and S. Siksek, Classical and modular approaches to exponential

Diophantine equations I. Fibonacci and Lucas perfect powers, Ann. of Math. (2), to appear.BS01 Y. Bugeaud and T. N. Shorey, On the number of solutions of the generalized Ramanujan–Nagell

equation, J. reine angew. Math. 539 (2001), 55–74.CF96 J. W. S. Cassels and E. V. Flynn, Prolegomena to a Middlebrow Arithmetic of Curves of Genus 2,

London Mathematical Society Lecture Note Series, vol. 230 (Cambridge University Press, Cam-bridge, 1996).

Coh93a H. Cohen, A Course in Computational Algebraic Number Theory, Graduate Text in Mathematics,vol. 138 (Springer, Berlin, 1993).

Coh93b J. H. E. Cohn, The Diophantine equation x2 + C = yn, Acta Arith. 55 (1993), 367–381.

60



https://doi.org/10.1112/S0010437X05001739



Coh03 J. H. E. Cohn, The Diophantine equation x2 + C = yn, II, Acta Arith. 109 (2003), 205–206.Cre96 J. E. Cremona, Algorithms for modular elliptic curves, second edition (Cambridge University

Press, Cambridge, 1996).Eul70 L. Euler, Vollstandige Einleitung zur Algebra, vol. 2 (St. Petersburg, 1770).GPZ98 J. Gebel, A. Petho and H. G. Zimmer, On Mordell’s equation, Compositio Math. 110 (1998),

335–367.Gou02 N. Gouillon, Un lemme de zeros, C. R. Acad. Sci. Paris Ser. I Math. 335 (2002), 167–170.Han00 G. Hanrot, Solving Thue equations without the full unit group, Math. Comp. 69 (2000), 395–405.Ivo03 W. Ivorra, Sur les equations xp +2βyp = z2 et xp +2βyp = 2z2, Acta Arith. 108 (2003), 327–338.Ko65 C. Ko, On the diophantine equation x2 = yn + 1, xy �= 0, Sci. Sinica (Notes) 14 (1965), 457–460.Kra98 A. Kraus, Sur l’equation a3 + b3 = cp, Experiment. Math. 7 (1998), 1–13.KO92 A. Kraus and J. Oesterle, Sur une question de B. Mazur, Math. Ann. 293 (1992), 259–275.Lau03 M. Laurent, Personal communication, November 2003.LM95 M. Laurent, M. Mignotte and Yu. Nesterenko, Formes lineares en deux logarithmes et

determinants d’interpolation, J. Number Theory 55 (1995), 255–265.Le02 M. Le, On Cohn’s conjecture concerning the Diophantine equation x2 + 2m = yn, Arch. Math.

(Basel) 78 (2002), 26–35.Leb50 V. A. Lebesgue, Sur l’impossibilite en nombres entiers de l’equation xm = y2 +1, Nouvelles Ann.

des Math. (1) 9 (1850), 178–181.Les98 J.-L. Lesage, Difference entre puissances et carres d’entiers, J. Number Theory 73 (1998),

390–425.Lju63 W. Ljunggren, On the diophantine equation y2 − k = x3, Acta Arith. 8 (1963), 451–463.Lju64 W. Ljunggren, On the diophantine equation Cx2 + D = yn, Pacific J. Math. 14 (1964), 585–596.Mat00 E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of

algebraic numbers. II, Izv. Ross. Akad. Nauk Ser. Mat. 64 (2000), 125–180. (English transl. Izv.Math. 64 (2000), 1217–1269.)

Maz78 B. Mazur, Rational isogenies of prime degree, Invent. Math. 44 (1978), 129–162.Mig84 M. Mignotte, Une nouvelle resolution de l’equation x2 + 7 = 2n, Rend. Sem. Fac. Sci. Univ.

Cagliari, 54 (1984), 41–43.Mig M. Mignotte, A kit on linear forms in three logarithms, Publ. IRMA, Strasbourg, to appear.MW96 M. Mignotte and B. M. M. de Weger, On the Diophantine equations x2+74 = y5 and x2+86 = y5,

Glasg. Math. J. 38 (1996), 77–85.Nag48 T. Nagell, Løsning til oppgave nr 2, 1943, s. 29, Nordisk Mat. Tidskr. 30 (1948), 62–64.Nag02 T. Nagell, Collected papers of Trygve Nagell, vols. 1–4, ed. P. Ribenboim, Queen’s Papers in Pure

and Applied Mathematics, vol. 121 (Queen’s University, Kingston, ON, 2002).PS97 B. Poonen and E. F. Schaefer, Explicit descent on cyclic covers of the projective line, J. reine

angew. Math. 488 (1997), 141–188.Ram13 S. Ramanujan, Question 464, J. Indian Math. Soc. 5 (1913), 120.Rib90 K. Ribet, On modular representations of Gal(Q/Q) arising from modular forms, Invent. Math.

100 (1990), 431–476.Sch95a E. F. Schaefer, 2-descent on the Jacobians of hyperelliptic curves, J. Number Theory 51 (1995),

219–232.Sch95b R. Schoof, Counting points on elliptic curves over finite fields, J. Theor. Nombres Bordeaux 7

(1995), 219–254.Ser87 J.-P. Serre, Sur les representations modulaires de degre 2 de Gal(Q/Q), Duke Math. J. 54 (1987),

179–230.Sik03 S. Siksek, On the diophantine equation x2 = yp + 2kzp, J. Theor. Nombres Bordeaux 15 (2003),

839–846.

61



https://doi.org/10.1112/S0010437X05001739



Sik S. Siksek, The modular approach to Diophantine equations, in Explicit Methods in Number Theory,Panoramas et Syntheses (Societe Mathematique de France), to appear.

SC03 S. Siksek and J. E. Cremona, On the Diophantine equation x2 +7 = ym, Acta Arith. 109 (2003),143–149.

Sma98 N. Smart, The Algorithmic Resolution of Diophantine Equations, London Mathematical SocietyStudent Texts, vol. 41 (Cambridge University Press, Cambridge, 1998).

Ste05a W. A. Stein, Modular Forms Database, http://modular.ucsd.edu/Tables.Ste05b W. A. Stein, An introduction to computing modular forms using modular symbols, in Algorithmic

Number Theory, eds J. Buhler and P. Stevenhagen (Cambridge University Press, Cambridge, toappear).

Sto98 M. Stoll, On the arithmetic of the curves y2 = xl + A and their Jacobians, J. reine angew. Math.501 (1998), 171–189.

Sto01 M. Stoll, Implementing 2-descent for Jacobians of hyperelliptic curves, Acta Arith. 98 (2001),245–277.

Sto02 M. Stoll, On the arithmetic of the curves y2 = xl +A, II, J. Number Theory 93 (2002), 183–206.TW95 R. L. Taylor and A. Wiles, Ring theoretic properties of certain Hecke algebras, Ann. of Math. (2)

141 (1995), 553–572.Wil95 A. Wiles, Modular elliptic curves and Fermat’s Last Theorem, Ann. of Math. (2) 141 (1995),

443–551.

Yann Bugeaud [email protected] Louis Pasteur, U. F. R. de mathematiques, 7, rue Rene Descartes, 67084 Strasbourgcedex, France

Maurice Mignotte [email protected] Louis Pasteur, U. F. R. de mathematiques, 7, rue Rene Descartes, 67084 Strasbourgcedex, France

Samir Siksek [email protected] of Mathematics, University of Warwick, Coventry, CV4 7AL, UK

62


mailto:[email protected]




https://doi.org/10.1112/S0010437X05001739


classical and modular approaches to exponential diophantine … · classical and modular approaches...

Documents